Physics - EMI/AC Question with Solution | TestHub

P: The self-inductance of a long solenoid is $L={\mu }_0{n}^{2}A{L}_s={\mu }_0(N_s/L_s{)}^2(\pi R_s^2)L_s=\frac{{\mu }_0{N}_s^2\pi R_s^2}{L_s}$. Q: The magnetic field inside the solenoid is $B={\mu }_0(N_s/L_s)I$. The flux through the smaller coil is ${\Phi }_c=N_{c}B(\pi R_c^2)=N_c{\mu }_0(N_s/L_s)I(\pi R_c^2)$. So, mutual inductance $M={\Phi }_c/I=\frac{{\mu }_0{N}_s{N}_c\pi R_c^2}{L_s}$. R: The induced EMF in the smaller coil is ${\mathcal{E}}_c=-M\frac{dI}{dt}=-M{I}_0\omega \cos (\omega t)$. Its peak value is $M{I}_0\omega =\frac{{\mu }_0{N}_s{N}_c\pi R_c^2{I}_0\omega }{L_s}$. S: The energy stored in the solenoid is $U=\frac{1}{2}L{I}_0^2=\frac{1}{2}(\frac{{\mu }_0{N}_s^2\pi R_s^2}{L_s})I_0^2$.